91Ó°ÊÓ

In the context of this exercise, we are dealing with a rectangular area that must be fenced. A rectangular area is defined by its length and width. The total area is calculated by multiplying these two dimensions together. In mathematical terms, if we let the length of the rectangle be represented as \(x\) and the width as \(y\), the area formula becomes \(x \cdot y\).

The problem specifies that the enclosed area measures 6000 square feet. This gives us the equation \(x \cdot y = 6000\). Understanding this is crucial because it links the dimensions of the rectangle to its total area, providing a basis for further calculations. By knowing the area and one dimension, you can easily compute the other dimension, which is an essential step in solving area problems.

Fencing this rectangular region involves different costs for different sides, adding another layer of complexity to the problem. Each side has a cost associated with fencing per linear foot. In this case, three sides cost \(\\( 3.75\) per foot, while the fourth side costs \(\\) 2.00\) per foot.

Cost per foot is a straightforward concept but vital for calculating total costs. Here's how you think about it: multiply the length of each side by its respective cost per foot to determine the expense for that side.

• For three sides at \(\\( 3.75\) per foot: it involves the width \(y\) and the length \(x\).
• For the fourth side at \(\\) 2.00\) per foot: it involves the length \(x\).

These values are then summed to determine the total cost of fencing the area.

Crafting an algebraic expression to depict the cost of fencing brings clarity to the problem. In algebra, expressions convey relationships and solve for unknowns using variables. For this scenario, the cost function \(C(x)\) provides the total fencing cost based on the length \(x\) of the fourth side. The formula is obtained by combining the costs of all sides. The resulting expression is: \(C(x) = 3.75(x + 2y) + 2.00x\).

To refine this cost function, you substitute \(y\) from the earlier area equation, namely: \(y = \frac{6000}{x}\). Insert this definition into \(C(x)\) to make it entirely dependent on \(x\), simplifying our equation to \(C(x) = 5.75x + \frac{45000}{x}\). This formula represents the functional relationship between the side length \(x\) and total fencing cost, factoring in all variable costs.

• Use of literals \(x\) and \(y\) to represent unknowns.
• Simplification by substituting known relationships like \(y = \frac{6000}{x}\).

Understanding this expression helps us make better decisions about dimensions and costs.

The challenge here is to solve an optimization problem, particularly about minimizing or understanding the most favorable total fence cost for the given rectangular configuration. An optimization problem seeks the best solution within defined constraints, such as area requirements or cost limits.

In this exercise, you explored how the choice of length \(x\) impacts the total cost \(C(x)\). The cost function \(C(x) = 5.75x + \frac{45000}{x}\) enables analysis of how costs evolve as dimensions change. The goal is to find a value for \(x\) that minimizes the cost while still enclosing the area of 6000 square feet.
Utilizing differential calculus or graphical analysis can aid in identifying the point at which the cost function reaches a minimum. This approach supports decisions about how best to allocate resources given certain constraints, a critical skill in real-world applications.

• Assessment of total cost implications based on varied lengths \(x\).
• Considerations of how physical constraints like area and cost drive decisions.

Through this problem, one learns about leveraging mathematical tools to optimize outcomes.